
\prob{00A3}{巧解四次方程III}

解方程
\[ \left(x^2 - \sqrt3\right)^2 = x + \sqrt3 \]
\problabels{yellow/代数, green/方程相关问题}

\ans{
  \begin{align*}
    x_{1,2} &= \frac12\left(-1 \pm \sqrt{4\sqrt3 - 3}\right) \\
    x_{3,4} &= \frac12\left(1 \pm \sqrt{4\sqrt3 + 1}\right)
  \end{align*}
}

\subsection{换常数为元}

原方程展开为
\[ x^4 - 2\sqrt3x^2 - x + 3 - \sqrt3 = 0 \]
令$k = \sqrt3$，则
\[ x^4 - 2x^2k - x + k^2 - k = 0 \]
于是得一关于$k$的方程
\[ k^2 - \left(2x^2 + 1\right)k + \left(x^4 - x\right) = 0 \]
解得
\begin{align*}
  k &= \frac12\left(2x^2 + 1 \pm \sqrt{\left(2x^2 + 1\right)^2 - 4\left(x^4 - x\right)}\right) \\
  &= \frac12\left(2x^2 + 1 \pm \sqrt{4x^2 + 4x + 1}\right) \\
  &= x^2 + \frac12 \pm \left(x + \frac12\right)
\end{align*}
故$k = x^2 + x + 1$或$k = x^2 - x$。而$k = \sqrt3$，解两关于$x$的二次方程得
\begin{align*}
  x_{1,2} &= \frac12\left(-1 \pm \sqrt{4\sqrt3 - 3}\right) \\
  x_{3,4} &= \frac12\left(1 \pm \sqrt{4\sqrt3 + 1}\right)
\end{align*}